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#1
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Polinomial graph
I have the following equation on my graph's trendline.
y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
#2
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Close, but not quite. The formula is y=ax^2+bx+c not (ax)^2+(bx)+c.
So, you want (0.0148*2^2)+(0.0375*2)-0.0398. Actually, the brackets are also not necessary because XL's operator precedence ranks exponentiation above multiplication above addition. Hence, 0.0148*2^2+0.0375*2-0.0398 is the same as the above equation. And, of course, if the value of x were in some cell, say, A4, then one would enter, in some other cell, the formula =0.0148*A4^2+0.0375*A4-0.0398 One final note. The trendline equation is not very accurate unless you increase the number of digits shown. Double-click the equation in the chart. Alternatively, use the LINEST function. Contrary to MS's documentation, it works for polynomials. See Bernard Liengme's http://www.stfx.ca/people/bliengme/E...Polynomial.htm Far more than you ever wanted to know about polynomial equations and XL, eh? {grin} -- Regards, Tushar Mehta www.tushar-mehta.com Excel, PowerPoint, and VBA add-ins, tutorials Custom MS Office productivity solutions In article , says... I have the following equation on my graph's trendline. y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
#3
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Hugh,
That looks correct. Another useful check would be to substitute 5 or 6 different points for X. Then chart the line. It should look exactly like your built-in trendline. For example, you are already substituting 2 for X below. In cell A1 add the number 1, in cell A2 add 2, in cell A3 add 3, in cell A4 add 4, an so on. Then in cell B1 reference your equation. Copy the equation down the range of column B to match the points in column A. Then chart column A as the X axis and column B as the Y axis. The charted trendline should look just like your built-in trendline. A B 1 =(0.0148(A1)^2)+0.0375(A1)-0.0398 2 =(0.0148(A2)^2)+0.0375(A2)-0.0398 3 =(0.0148(A3)^2)+0.0375(A3)-0.0398 4 =(0.0148(A4)^2)+0.0375(A4)-0.0398 5 =(0.0148(A5)^2)+0.0375(A5)-0.0398 6 =(0.0148(A6)^2)+0.0375(A6)-0.0398 ---- Regards, John Mansfield http://www.pdbook.com "Hugh" wrote: I have the following equation on my graph's trendline. y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
#4
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Many thanks John,
I don't seem to get the same curve when I use the actual x, as opposed to the 1,2,3 test figures, hence the belief that I must have had the formula wrong. I'll try it again and see if I can get a fit. Are there any rules that that could be throwing my graph off, like 0s, or too high x numbers? Hugh "John Mansfield" wrote in message ... Hugh, That looks correct. Another useful check would be to substitute 5 or 6 different points for X. Then chart the line. It should look exactly like your built-in trendline. For example, you are already substituting 2 for X below. In cell A1 add the number 1, in cell A2 add 2, in cell A3 add 3, in cell A4 add 4, an so on. Then in cell B1 reference your equation. Copy the equation down the range of column B to match the points in column A. Then chart column A as the X axis and column B as the Y axis. The charted trendline should look just like your built-in trendline. A B 1 =(0.0148(A1)^2)+0.0375(A1)-0.0398 2 =(0.0148(A2)^2)+0.0375(A2)-0.0398 3 =(0.0148(A3)^2)+0.0375(A3)-0.0398 4 =(0.0148(A4)^2)+0.0375(A4)-0.0398 5 =(0.0148(A5)^2)+0.0375(A5)-0.0398 6 =(0.0148(A6)^2)+0.0375(A6)-0.0398 ---- Regards, John Mansfield http://www.pdbook.com "Hugh" wrote: I have the following equation on my graph's trendline. y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
#5
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Hugh -
Also, use an XY (Scatter) chart type, not a Line chart type. - Mike www.mikemiddleton.com "Hugh" wrote in message ... Many thanks John, I don't seem to get the same curve when I use the actual x, as opposed to the 1,2,3 test figures, hence the belief that I must have had the formula wrong. I'll try it again and see if I can get a fit. Are there any rules that that could be throwing my graph off, like 0s, or too high x numbers? Hugh "John Mansfield" wrote in message ... Hugh, That looks correct. Another useful check would be to substitute 5 or 6 different points for X. Then chart the line. It should look exactly like your built-in trendline. For example, you are already substituting 2 for X below. In cell A1 add the number 1, in cell A2 add 2, in cell A3 add 3, in cell A4 add 4, an so on. Then in cell B1 reference your equation. Copy the equation down the range of column B to match the points in column A. Then chart column A as the X axis and column B as the Y axis. The charted trendline should look just like your built-in trendline. A B 1 =(0.0148(A1)^2)+0.0375(A1)-0.0398 2 =(0.0148(A2)^2)+0.0375(A2)-0.0398 3 =(0.0148(A3)^2)+0.0375(A3)-0.0398 4 =(0.0148(A4)^2)+0.0375(A4)-0.0398 5 =(0.0148(A5)^2)+0.0375(A5)-0.0398 6 =(0.0148(A6)^2)+0.0375(A6)-0.0398 ---- Regards, John Mansfield http://www.pdbook.com "Hugh" wrote: I have the following equation on my graph's trendline. y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
#6
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And follow Tushar's advice to display many more digits of the trendline
equation. - Jon ------- Jon Peltier, Microsoft Excel MVP Peltier Technical Services Tutorials and Custom Solutions http://PeltierTech.com/ _______ Michael R Middleton wrote: Hugh - Also, use an XY (Scatter) chart type, not a Line chart type. - Mike www.mikemiddleton.com "Hugh" wrote in message ... Many thanks John, I don't seem to get the same curve when I use the actual x, as opposed to the 1,2,3 test figures, hence the belief that I must have had the formula wrong. I'll try it again and see if I can get a fit. Are there any rules that that could be throwing my graph off, like 0s, or too high x numbers? Hugh "John Mansfield" wrote in message ... Hugh, That looks correct. Another useful check would be to substitute 5 or 6 different points for X. Then chart the line. It should look exactly like your built-in trendline. For example, you are already substituting 2 for X below. In cell A1 add the number 1, in cell A2 add 2, in cell A3 add 3, in cell A4 add 4, an so on. Then in cell B1 reference your equation. Copy the equation down the range of column B to match the points in column A. Then chart column A as the X axis and column B as the Y axis. The charted trendline should look just like your built-in trendline. A B 1 =(0.0148(A1)^2)+0.0375(A1)-0.0398 2 =(0.0148(A2)^2)+0.0375(A2)-0.0398 3 =(0.0148(A3)^2)+0.0375(A3)-0.0398 4 =(0.0148(A4)^2)+0.0375(A4)-0.0398 5 =(0.0148(A5)^2)+0.0375(A5)-0.0398 6 =(0.0148(A6)^2)+0.0375(A6)-0.0398 ---- Regards, John Mansfield http://www.pdbook.com "Hugh" wrote: I have the following equation on my graph's trendline. y=0.0148xsquared +0.0375x-0.0398. If x=2 would the formula for trend be ((0.0148*2)^2)+(0.0375*2)-0.0398? Thanks |
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